Write a short description about the course and add a link to your GitHub repository here. This is an R Markdown (.Rmd) file so you should use R Markdown syntax.
The course seems to be set up nicely. At least much better than some online courses about statistics/programming. I hope I’ll relearn to use R (briefly tried it about twelve years ago). RStudio seems convenient and I like the GitHub integration. Additionally, I hope to deepen my understanding of statistical methods.
My GitHub repository: https://github.com/MTurkkila/IODS-projectlink
I started with the Datacamp and afterwards th data wrangling was straightforward and easy. Thus I used the data file from my local folder in the analysis. Below are the important parts of the script. Before any analysis, ggplot2 and GGally libraries are accessed first as is the common practice.
library(ggplot2)
library(GGally)
Read the data from a local folder and check for structure and dimensions
students2014 <- read.table("data/students2014.txt", sep="\t")
str(students2014)
## 'data.frame': 166 obs. of 7 variables:
## $ gender : Factor w/ 2 levels "F","M": 1 2 1 2 2 1 2 1 2 1 ...
## $ Age : int 53 55 49 53 49 38 50 37 37 42 ...
## $ Attitude: int 37 31 25 35 37 38 35 29 38 21 ...
## $ deep : num 3.58 2.92 3.5 3.5 3.67 ...
## $ stra : num 3.38 2.75 3.62 3.12 3.62 ...
## $ surf : num 2.58 3.17 2.25 2.25 2.83 ...
## $ Points : int 25 12 24 10 22 21 21 31 24 26 ...
dim(students2014)
## [1] 166 7
The Data has 7 variables and 166 observations from a questionnaire. Variables deep, stra and surf are sum variables scaled to the original scale. The data only includes observations with points over 0.
Next I use the ggpair-function to show graphical overview of the data. In the actual script, I also save the plots in a local folder plots with dev.copy()-function.
p <- ggpairs(students2014, mapping = aes(col = gender, alpha = 0.3), lower = list(combo = wrap("facethist", bins = 20)))
p
In addition to plotting the dataframe, I checked the summaries of the data.
summary(students2014)
## gender Age Attitude deep stra
## F:110 Min. :17.00 Min. :14.00 Min. :1.583 Min. :1.250
## M: 56 1st Qu.:21.00 1st Qu.:26.00 1st Qu.:3.333 1st Qu.:2.625
## Median :22.00 Median :32.00 Median :3.667 Median :3.188
## Mean :25.51 Mean :31.43 Mean :3.680 Mean :3.121
## 3rd Qu.:27.00 3rd Qu.:37.00 3rd Qu.:4.083 3rd Qu.:3.625
## Max. :55.00 Max. :50.00 Max. :4.917 Max. :5.000
## surf Points
## Min. :1.583 Min. : 7.00
## 1st Qu.:2.417 1st Qu.:19.00
## Median :2.833 Median :23.00
## Mean :2.787 Mean :22.72
## 3rd Qu.:3.167 3rd Qu.:27.75
## Max. :4.333 Max. :33.00
As attitude, stra and surf have the highest absolute correlation with points I chose those for the initial linear model and created a regression model with multiple explanatory variables and print the summary of my model.
my_model <- lm(Points ~ Attitude + stra + surf, data = students2014)
summary(my_model)
##
## Call:
## lm(formula = Points ~ Attitude + stra + surf, data = students2014)
##
## Residuals:
## Min 1Q Median 3Q Max
## -17.1550 -3.4346 0.5156 3.6401 10.8952
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 11.01711 3.68375 2.991 0.00322 **
## Attitude 0.33952 0.05741 5.913 1.93e-08 ***
## stra 0.85313 0.54159 1.575 0.11716
## surf -0.58607 0.80138 -0.731 0.46563
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 5.296 on 162 degrees of freedom
## Multiple R-squared: 0.2074, Adjusted R-squared: 0.1927
## F-statistic: 14.13 on 3 and 162 DF, p-value: 3.156e-08
The summary shows estimations for the parameters of the linear model and the statistical significance for those estimates. Only attitude is statistically significant (shown with the stars) and therefore I only include it to the second model
my_model2 <- lm(Points ~ Attitude, data = students2014)
summary(my_model2)
##
## Call:
## lm(formula = Points ~ Attitude, data = students2014)
##
## Residuals:
## Min 1Q Median 3Q Max
## -16.9763 -3.2119 0.4339 4.1534 10.6645
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 11.63715 1.83035 6.358 1.95e-09 ***
## Attitude 0.35255 0.05674 6.214 4.12e-09 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 5.32 on 164 degrees of freedom
## Multiple R-squared: 0.1906, Adjusted R-squared: 0.1856
## F-statistic: 38.61 on 1 and 164 DF, p-value: 4.119e-09
The summaries show that now both \(\alpha\) and \(\beta\) variables are highly statistically significant. The estimates show that attitude has positive correlation with the attitude i.e. more positive attitude yields more points.
The multiple R squared shows how much of the variation in the points is explains by the attitude. In this model, it is 19%
Check for validity with diagnostic plots
par(mfrow = c(2,2))
plot(my_model2, which = c(1, 2, 5))
The model assumes that errors are normally distributed, are not correlated and have constant variance.
Overall, I’d say the model is quite reasonable.
The data includes background information and alcohol consumption of Portuguese students. The data is joined from two original data sets of student performance in math and in Portuguese language. The joined data includes two new variables: alc_use and high_use. alc_use is mean of weekend alcohol use (Walc) and workday alcohol use (Dalc). Variable high_use is boolean value with condition: high_use > 2.
Below is list of the variables in the data. The complete description of original data can be found at UCI Machine Learning repository
alc <- read.csv("data/alc.csv")
variable.names(alc)
## [1] "school" "sex" "age" "address" "famsize"
## [6] "Pstatus" "Medu" "Fedu" "Mjob" "Fjob"
## [11] "reason" "nursery" "internet" "guardian" "traveltime"
## [16] "studytime" "failures" "schoolsup" "famsup" "paid"
## [21] "activities" "higher" "romantic" "famrel" "freetime"
## [26] "goout" "Dalc" "Walc" "health" "absences"
## [31] "G1" "G2" "G3" "alc_use" "high_use"
The purpose of this analysis is to examine relationship between alcohol consumption and free time. Four variables related to the free time and thus chosen for the analysis are:
The hypothesis is that more free time and going out a student has greater the chance for high alcohol consumption.
Firstly, to study distributions of the chosen variables, let’s draw bar plots:
library(tidyr); library(dplyr); library(ggplot2)
##
## Attaching package: 'dplyr'
## The following objects are masked from 'package:stats':
##
## filter, lag
## The following objects are masked from 'package:base':
##
## intersect, setdiff, setequal, union
alc4 <- c("freetime", "traveltime", "goout","activities")
gather(alc[alc4]) %>% ggplot(aes(value)) + facet_wrap("key", scales = "free") + geom_bar()
Free time and going out seems to be quite normally distributed. Home to school travel time is heavily skewed to short, under 15 minutes travel times. Extra-curricular activities is split quite evenly. Thus, it can be assumed that variables free time and going out has more significance in alcohol consumption.
Let’s also do simple cross-tabulation of high alcohol consumption with median of free time and going out.
alc %>% group_by(high_use) %>% summarise(count = n(), going_out = median(goout), freetime = median(freetime))
## `summarise()` ungrouping output (override with `.groups` argument)
## # A tibble: 2 x 4
## high_use count going_out freetime
## <lgl> <int> <dbl> <dbl>
## 1 FALSE 268 3 3
## 2 TRUE 114 4 3
Median for going out is higher for the group that goes out more. It might be that free time in itself does not affect alcohol consumption as much. However let’s start the logistic model with all four variables as is instructed.
To build the logistic model I use glm-function with all four variables and print the summary of the model
m <- glm(high_use ~ freetime + traveltime + goout + activities, data = alc, family = "binomial")
summary(m)
##
## Call:
## glm(formula = high_use ~ freetime + traveltime + goout + activities,
## family = "binomial", data = alc)
##
## Deviance Residuals:
## Min 1Q Median 3Q Max
## -1.5791 -0.7737 -0.5840 0.9795 2.3555
##
## Coefficients:
## Estimate Std. Error z value Pr(>|z|)
## (Intercept) -4.1598 0.6076 -6.847 7.56e-12 ***
## freetime 0.1524 0.1340 1.137 0.2553
## traveltime 0.4226 0.1723 2.452 0.0142 *
## goout 0.7225 0.1213 5.959 2.54e-09 ***
## activitiesyes -0.3633 0.2446 -1.485 0.1376
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## (Dispersion parameter for binomial family taken to be 1)
##
## Null deviance: 465.68 on 381 degrees of freedom
## Residual deviance: 406.48 on 377 degrees of freedom
## AIC: 416.48
##
## Number of Fisher Scoring iterations: 4
As expected, going out is significant predictor for high alcohol consumption. Surprisingly, also travel time has statistical signifigance. Let’s build new model with these two variables.
m <- glm(high_use ~ traveltime + goout, data = alc, family = "binomial")
summary(m)
##
## Call:
## glm(formula = high_use ~ traveltime + goout, family = "binomial",
## data = alc)
##
## Deviance Residuals:
## Min 1Q Median 3Q Max
## -1.4676 -0.8436 -0.6050 1.0674 2.3813
##
## Coefficients:
## Estimate Std. Error z value Pr(>|z|)
## (Intercept) -3.9443 0.4973 -7.932 2.16e-15 ***
## traveltime 0.4142 0.1712 2.419 0.0156 *
## goout 0.7553 0.1165 6.485 8.87e-11 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## (Dispersion parameter for binomial family taken to be 1)
##
## Null deviance: 465.68 on 381 degrees of freedom
## Residual deviance: 409.76 on 379 degrees of freedom
## AIC: 415.76
##
## Number of Fisher Scoring iterations: 4
Compute odds ratios OR and confidence intervals CI and print them.
OR <- coef(m) %>% exp
CI <- confint(m) %>% exp
cbind(OR, CI)
## OR 2.5 % 97.5 %
## (Intercept) 0.01936554 0.007015459 0.04951509
## traveltime 1.51309541 1.083767881 2.12550643
## goout 2.12821744 1.703804423 2.69227727
As the odds ratio is higher that 1 for both variables, it means that both variables are positively associated with the high alcohol consumption.
To explore the predictive power of the logistic model m, I use predict() function to make predictions for high alcohol consumption. The probabilities from the predictions are added to the alc dataframe as well as boolean value with the same condition, >0.5 as the observational data. Finally we print cross tabulation of predictions and actual values.
probabilities <- predict(m, type = "response")
alc <- mutate(alc, probability = probabilities)
alc <- mutate(alc, prediction = probability > 0.5)
table(high_use = alc$high_use, prediction = alc$prediction)
## prediction
## high_use FALSE TRUE
## FALSE 247 21
## TRUE 78 36
Let’s also plot the actual values and the predictions
g <- ggplot(alc, aes(x = probability, y = high_use, col = prediction))
g + geom_point()
Looks like there are quite many false predictions so let’s check the training error by first defining loss function
loss_func <- function(class, prob) {
n_wrong <- abs(class - prob) > 0.5
mean(n_wrong)
}
Next we call that loss function to compute percentage of wrong predictions
loss_func(class = alc$high_use, prob = alc$probability)
## [1] 0.2591623
The result shows that around 26% of the predictions are wrong. The model is not very good at predictions, but it still is somewhat better than simply fifty-fifty guessing.
Let’s use cv.glm() function form ‘boot’ library for K-fold cross-validation and start with 10-fold cross-validation. The function returns vector ‘delta’ which first component is the estimate for prediction error. The cv.glm() function uses the previously defined loss functions as the cost function.
library(boot)
cv <- cv.glm(data = alc, cost = loss_func, glmfit = m, K = 10)
cv$delta[1]
## [1] 0.2591623
The predictor error is slightly higher than the training error and it is also the close to the error in the datacamp exercise. Let’s check if it is possible to find logistic model with smaller error.
The next bit of code uses ten different logistic regression models with different number of predictor variables. In a for loop prediction probabilities are computed for each model. Additionally, 10-fold cross-validation is performed. Finally, training and prediction errors are plotted against number of predictors
# Dataframe for training and prediction errors
errors <- data.frame(matrix(ncol=3,nrow=0, dimnames=list(NULL, c("n_pred", "training", "prediction"))))
# Models with different number of predictors
m1 <- high_use ~ goout + traveltime + freetime + activities + studytime + paid + romantic+G1+G2+G3
m2 <- high_use ~ goout + traveltime + freetime + activities + studytime + paid + romantic+G1+G2
m3 <- high_use ~ goout + traveltime + freetime + activities + studytime + paid + romantic+G1
m4 <- high_use ~ goout + traveltime + freetime + activities + studytime + paid + romantic
m5 <- high_use ~ goout + traveltime + freetime + activities + studytime + paid
m6 <- high_use ~ goout + traveltime + freetime + activities + studytime
m7 <- high_use ~ goout + traveltime + freetime + activities # The original model
m8 <- high_use ~ goout + traveltime + freetime
m9 <- high_use ~ goout + traveltime # The updated model used in assignments
m10 <- high_use ~ goout
for(i in 1:10) {
tmp <- paste0("m",i)
m <- glm(tmp, data = alc, family = "binomial")
probabilities <- predict(m, type = "response")
alc <- mutate(alc, probability = probabilities)
alc <- mutate(alc, prediction = probability > 0.5)
cv <- cv.glm(data = alc, cost = loss_func, glmfit = m, K = 10)
errors[i, "n_pred"] <- 11-i
errors[i, "training"] <- loss_func(class = alc$high_use, prob = alc$probability)
errors[i, "prediction"] <- cv$delta[1]
}
g <- errors %>% gather(key,error, training, prediction) %>%ggplot(aes(x=n_pred, y=error, colour=key))
g + geom_line() + scale_x_discrete(limits=c(1:10), name = "Number of predictors")
Both errors decrease slightly when number of predictors increases, but the trend is not linear or even stable. Also the absolute amount does not change that much only from 26% to 22% in training error between 2 (the model used in th exercises) and 6 predictors. If I check the summary of that six predictor model, only studytime has statistical significance in addition to the 2 predictor model. Of course, it would be possible to continue with this exploration and maybe even find smaller training and prediction error. Then we would be close to machine learning and that’s a topic for another time.
m <- glm(m5, data = alc, family = "binomial")
summary(m)
##
## Call:
## glm(formula = m5, family = "binomial", data = alc)
##
## Deviance Residuals:
## Min 1Q Median 3Q Max
## -1.6671 -0.7784 -0.5273 0.8741 2.4554
##
## Coefficients:
## Estimate Std. Error z value Pr(>|z|)
## (Intercept) -3.1062 0.7143 -4.348 1.37e-05 ***
## goout 0.7234 0.1218 5.941 2.84e-09 ***
## traveltime 0.3829 0.1754 2.183 0.029070 *
## freetime 0.1199 0.1370 0.875 0.381614
## activitiesyes -0.2689 0.2513 -1.070 0.284700
## studytime -0.5986 0.1717 -3.487 0.000488 ***
## paidyes 0.4775 0.2549 1.873 0.061055 .
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## (Dispersion parameter for binomial family taken to be 1)
##
## Null deviance: 465.68 on 381 degrees of freedom
## Residual deviance: 391.34 on 375 degrees of freedom
## AIC: 405.34
##
## Number of Fisher Scoring iterations: 4
In this chapter I look into linear discriminant analysis and k-means clustering using ready made data set from the MASS-library. please also see bonus, tho not ready yet.
Access the required libraries
library(MASS); library(tidyr); library(ggplot2); library(corrplot)
First I load the Boston data set and explore it with str and summary.
data("Boston")
str(Boston)
## 'data.frame': 506 obs. of 14 variables:
## $ crim : num 0.00632 0.02731 0.02729 0.03237 0.06905 ...
## $ zn : num 18 0 0 0 0 0 12.5 12.5 12.5 12.5 ...
## $ indus : num 2.31 7.07 7.07 2.18 2.18 2.18 7.87 7.87 7.87 7.87 ...
## $ chas : int 0 0 0 0 0 0 0 0 0 0 ...
## $ nox : num 0.538 0.469 0.469 0.458 0.458 0.458 0.524 0.524 0.524 0.524 ...
## $ rm : num 6.58 6.42 7.18 7 7.15 ...
## $ age : num 65.2 78.9 61.1 45.8 54.2 58.7 66.6 96.1 100 85.9 ...
## $ dis : num 4.09 4.97 4.97 6.06 6.06 ...
## $ rad : int 1 2 2 3 3 3 5 5 5 5 ...
## $ tax : num 296 242 242 222 222 222 311 311 311 311 ...
## $ ptratio: num 15.3 17.8 17.8 18.7 18.7 18.7 15.2 15.2 15.2 15.2 ...
## $ black : num 397 397 393 395 397 ...
## $ lstat : num 4.98 9.14 4.03 2.94 5.33 ...
## $ medv : num 24 21.6 34.7 33.4 36.2 28.7 22.9 27.1 16.5 18.9 ...
summary(Boston)
## crim zn indus chas
## Min. : 0.00632 Min. : 0.00 Min. : 0.46 Min. :0.00000
## 1st Qu.: 0.08204 1st Qu.: 0.00 1st Qu.: 5.19 1st Qu.:0.00000
## Median : 0.25651 Median : 0.00 Median : 9.69 Median :0.00000
## Mean : 3.61352 Mean : 11.36 Mean :11.14 Mean :0.06917
## 3rd Qu.: 3.67708 3rd Qu.: 12.50 3rd Qu.:18.10 3rd Qu.:0.00000
## Max. :88.97620 Max. :100.00 Max. :27.74 Max. :1.00000
## nox rm age dis
## Min. :0.3850 Min. :3.561 Min. : 2.90 Min. : 1.130
## 1st Qu.:0.4490 1st Qu.:5.886 1st Qu.: 45.02 1st Qu.: 2.100
## Median :0.5380 Median :6.208 Median : 77.50 Median : 3.207
## Mean :0.5547 Mean :6.285 Mean : 68.57 Mean : 3.795
## 3rd Qu.:0.6240 3rd Qu.:6.623 3rd Qu.: 94.08 3rd Qu.: 5.188
## Max. :0.8710 Max. :8.780 Max. :100.00 Max. :12.127
## rad tax ptratio black
## Min. : 1.000 Min. :187.0 Min. :12.60 Min. : 0.32
## 1st Qu.: 4.000 1st Qu.:279.0 1st Qu.:17.40 1st Qu.:375.38
## Median : 5.000 Median :330.0 Median :19.05 Median :391.44
## Mean : 9.549 Mean :408.2 Mean :18.46 Mean :356.67
## 3rd Qu.:24.000 3rd Qu.:666.0 3rd Qu.:20.20 3rd Qu.:396.23
## Max. :24.000 Max. :711.0 Max. :22.00 Max. :396.90
## lstat medv
## Min. : 1.73 Min. : 5.00
## 1st Qu.: 6.95 1st Qu.:17.02
## Median :11.36 Median :21.20
## Mean :12.65 Mean :22.53
## 3rd Qu.:16.95 3rd Qu.:25.00
## Max. :37.97 Max. :50.00
The MASS package contains data sets to accompany book “Modern Applied Statistics with S” by W. N. Venables and B. D. Ripley with several distinct data from different sources. The Boston data set contains Housing values in suburbs of Boston. The set includes, for example, variables crim that is the per capita crime rate by town and age that is proportion of owner-occupied units built prior to 1940. For complete descriptions please see (Boston {MASS})[https://stat.ethz.ch/R-manual/R-devel/library/MASS/html/Boston.html].
Next, let’s plot distributions of the variables. For fun, let’s use color of the Faculty of Educational Sciences.
gather(Boston) %>% ggplot(aes(value)) + facet_wrap("key", scales = "free") + geom_bar(fill="#fcd116", color="#fcd116", alpha=0.9)
From the bar charts it is apparent that the data is not normally distributed and, in most cases, strongly skewed.
To asses different relationships of the variables let’s plot correlation matrix. I think lower triangular matrix looks better so let’s use type = "lower" instead upper as it was in the datacamp exercises.
cor_matrix <-cor (Boston)
cor_matrix <- cor_matrix %>% round(2)
corrplot(cor_matrix, method="circle", type="lower", cl.pos = "b", tl.pos = "d", tl.cex = 0.6)
The per capita crime rate most strongly correlates positively with variables
rad and tax that are index of accessibility to radial highways and full-value property-tax rate per $10,000 respectively. These two variables are also strongly correlated together. I can not reason for the connection between these variables ans the crime rate.
For the following analyzes we need to standardize the data I use the scale function. The function subtracts column mean from each value and then divides it with the standard deviation: \(scaled(x) = \frac{x-\bar{x}}{\mu_x}\)
We can see this effect with the summary.
boston_scaled <- scale(Boston)
summary(boston_scaled)
## crim zn indus chas
## Min. :-0.419367 Min. :-0.48724 Min. :-1.5563 Min. :-0.2723
## 1st Qu.:-0.410563 1st Qu.:-0.48724 1st Qu.:-0.8668 1st Qu.:-0.2723
## Median :-0.390280 Median :-0.48724 Median :-0.2109 Median :-0.2723
## Mean : 0.000000 Mean : 0.00000 Mean : 0.0000 Mean : 0.0000
## 3rd Qu.: 0.007389 3rd Qu.: 0.04872 3rd Qu.: 1.0150 3rd Qu.:-0.2723
## Max. : 9.924110 Max. : 3.80047 Max. : 2.4202 Max. : 3.6648
## nox rm age dis
## Min. :-1.4644 Min. :-3.8764 Min. :-2.3331 Min. :-1.2658
## 1st Qu.:-0.9121 1st Qu.:-0.5681 1st Qu.:-0.8366 1st Qu.:-0.8049
## Median :-0.1441 Median :-0.1084 Median : 0.3171 Median :-0.2790
## Mean : 0.0000 Mean : 0.0000 Mean : 0.0000 Mean : 0.0000
## 3rd Qu.: 0.5981 3rd Qu.: 0.4823 3rd Qu.: 0.9059 3rd Qu.: 0.6617
## Max. : 2.7296 Max. : 3.5515 Max. : 1.1164 Max. : 3.9566
## rad tax ptratio black
## Min. :-0.9819 Min. :-1.3127 Min. :-2.7047 Min. :-3.9033
## 1st Qu.:-0.6373 1st Qu.:-0.7668 1st Qu.:-0.4876 1st Qu.: 0.2049
## Median :-0.5225 Median :-0.4642 Median : 0.2746 Median : 0.3808
## Mean : 0.0000 Mean : 0.0000 Mean : 0.0000 Mean : 0.0000
## 3rd Qu.: 1.6596 3rd Qu.: 1.5294 3rd Qu.: 0.8058 3rd Qu.: 0.4332
## Max. : 1.6596 Max. : 1.7964 Max. : 1.6372 Max. : 0.4406
## lstat medv
## Min. :-1.5296 Min. :-1.9063
## 1st Qu.:-0.7986 1st Qu.:-0.5989
## Median :-0.1811 Median :-0.1449
## Mean : 0.0000 Mean : 0.0000
## 3rd Qu.: 0.6024 3rd Qu.: 0.2683
## Max. : 3.5453 Max. : 2.9865
The scaling does not change the distribution of the data but scales all variable values around zero (mean).
The scaled data set is a matrix instead of dataframe, so let’s change it back. Replotting distributions, we see that the actual distributions do not change, but the values are
class(boston_scaled)
## [1] "matrix"
boston_scaled <- as.data.frame(boston_scaled)
Next, I create categorical variable of the crime rate by first dividing the data into four (quantile) bins and then cutting the data to those bins and giving each bin a label. Also, original crim variable is removed and the categorical variable added to the scaled data set.
bins <- quantile(boston_scaled$crim)
crime <- cut(boston_scaled$crim, breaks = bins, include.lowest = TRUE, labels = c("low", "med_low", "med_high", "high"))
boston_scaled <- dplyr::select(boston_scaled, -crim)
boston_scaled <- data.frame(boston_scaled, crime)
Additionally, I divide the data set to train and test sets for future use. To select 80% of the data randomly as the train set, I use sample to randomly select n*0.8 indexes from the scaled data set. Then, I use those index to select the data. Using [-ind,] selects the rows not in [ind,]. Lastly, I save the correct crime classes form test set and remove them from that set.
n <- nrow(boston_scaled)
ind <- sample(n, size = n * 0.8)
train <- boston_scaled[ind,]
test <- boston_scaled[-ind,]
correct_classes <- test$crime
test <- dplyr::select(test, -crime)
More about LDA in this StatQuest-video
For the linear discriminant analysis I use the categorical crime rate created above as the target variable and all other variables as predictors. Data for the fit is the train set previously cut from the boston_scaled dataframe. Next, I draw the LDA plot.
palette(c("#fcd116","#00a6ed","#f6511d", "#7fb800")) # Custom color palette
lda.fit <- lda(crime ~., data = train)
classes <- as.numeric(train$crime)
plot(lda.fit, dimen = 2, col = classes, pch = classes)
Now, it is possible to predict crime rate categories using the LDA fit. Let’s check predictions from the test set agains the correct classes and cross tabulate the results.
lda.pred <- predict(lda.fit, newdata = test)
table(correct = correct_classes, predicted = lda.pred$class)
## predicted
## correct low med_low med_high high
## low 6 14 1 0
## med_low 7 7 12 0
## med_high 1 14 12 1
## high 0 0 0 27
The predictions are quite good as most values are on the diagonal. Only med_low gets several false positives.
Before k-means clustering, I standardize the orginial Boston data set and calculate euclidean and manhattan distance.
Boston <- scale(Boston)
Boston <- as.data.frame(Boston)
dist_eu <- dist(Boston)
summary(dist_eu)
## Min. 1st Qu. Median Mean 3rd Qu. Max.
## 0.1343 3.4625 4.8241 4.9111 6.1863 14.3970
dist_man <- dist(Boston, method = "manhattan")
summary(dist_man)
## Min. 1st Qu. Median Mean 3rd Qu. Max.
## 0.2662 8.4832 12.6090 13.5488 17.7568 48.8618
The euclidean distances sre clearly shorter as, I think, they should be.
Next I do k-means clustering with three centers and draw correlation plots, with data point colors corresponding with clusters, of a few intresting variables.
km <-kmeans(Boston, centers = 3)
pairs(Boston[,c(1,9, 10, 12, 13, 14)], col = km$cluster)
To determine optimal number of clusters I calculate and visualize within cluster sum of squares (WCSS) as function of clusters. As kmeans assigns initial clusters randomly, I set constant seed used in random number generator.
set.seed(123) # the seed for rng
k_max <- 10
twcss <- sapply(1:k_max, function(k){kmeans(Boston, k)$tot.withinss}) # these lines form datacamp exercise
qplot(x = 1:k_max, y = twcss, geom = 'line') + scale_x_continuous(breaks=c(1:10))
The twcss drops most drastically from one to two clusters meaning that two clusters is the optimal number of clusters.
km <-kmeans(Boston, centers = 2)
pairs(Boston[,c(1,9, 10, 12, 13, 14)], col = km$cluster)
Visually also two clusters looks better than three.
For this bonus assignment I use LDA fit for three clusters from k-means algorithm.
Boston <- dplyr::select(Boston, -chas) # need remove this variable as the lda.fit doesn't work with it when knitting
km <-kmeans(Boston, centers = 3)
lda.fit <- lda(km$cluster ~., data = Boston)
Visualization of clusters and arrows done with code from datacamp.
lda.arrows <- function(x, myscale = 1, arrow_heads = 0.1, color = "red", tex = 0.75, choices = c(1,2)){
heads <- coef(x)
arrows(x0 = 0, y0 = 0,
x1 = myscale * heads[,choices[1]],
y1 = myscale * heads[,choices[2]], col=color, length = arrow_heads)
text(myscale * heads[,choices], labels = row.names(heads),
cex = tex, col=color, pos=3)
}
# target classes as numeric
classes <- as.numeric(km$cluster)
# plot the lda results
plot(lda.fit, dimen = 2, col = classes, pch = classes, )
lda.arrows(lda.fit, myscale = 3, color="black")
With three clusters it seem like variables age, zn, tax and nox are the most influential separators.
The code below for the (scaled) train data that you used to fit the LDA. The code creates a matrix product, which is a projection of the data points.
lda.fit <- lda(crime ~., data = train)
model_predictors <- dplyr::select(train, -crime)
# matrix multiplication
matrix_product <- as.matrix(model_predictors) %*% lda.fit$scaling
matrix_product <- as.data.frame(matrix_product)
library(plotly)
plot_ly(x = matrix_product$LD1, y = matrix_product$LD2, z = matrix_product$LD3, type= 'scatter3d', mode='markers', color = train$crime)
Access the libraries and load the human data.
library(GGally); library(dplyr); library(corrplot)
human <- read.csv("data/human2.csv", row.names = 1)
str(human)
## 'data.frame': 155 obs. of 8 variables:
## $ edu2_ratio: num 1.007 0.997 0.983 0.989 0.969 ...
## $ lab_ratio : num 0.891 0.819 0.825 0.884 0.829 ...
## $ Life_exp : num 81.6 82.4 83 80.2 81.6 80.9 80.9 79.1 82 81.8 ...
## $ Year_edu : num 17.5 20.2 15.8 18.7 17.9 16.5 18.6 16.5 15.9 19.2 ...
## $ GNI : int 64992 42261 56431 44025 45435 43919 39568 52947 42155 32689 ...
## $ MMR : int 4 6 6 5 6 7 9 28 11 8 ...
## $ ABR : num 7.8 12.1 1.9 5.1 6.2 3.8 8.2 31 14.5 25.3 ...
## $ Rep : num 39.6 30.5 28.5 38 36.9 36.9 19.9 19.4 28.2 31.4 ...
ggpairs(human) # miksi tulee nuo plot rivit?
cor(human) %>% corrplot() # muotoile kivemmaksi
pca_human <- prcomp(human)
s <- summary(pca_human)
pca_pr <- round(100*s$importance[2, ], digits = 1)
pc_lab <- paste0(names(pca_pr), " (", pca_pr, "%)")
biplot(pca_human, cex = c(0.8, 1), col = c("grey40", "deeppink2"), xlab = pc_lab[1], ylab = pc_lab[2])
## Warning in arrows(0, 0, y[, 1L] * 0.8, y[, 2L] * 0.8, col = col[2L], length =
## arrow.len): zero-length arrow is of indeterminate angle and so skipped
## Warning in arrows(0, 0, y[, 1L] * 0.8, y[, 2L] * 0.8, col = col[2L], length =
## arrow.len): zero-length arrow is of indeterminate angle and so skipped
## Warning in arrows(0, 0, y[, 1L] * 0.8, y[, 2L] * 0.8, col = col[2L], length =
## arrow.len): zero-length arrow is of indeterminate angle and so skipped
## Warning in arrows(0, 0, y[, 1L] * 0.8, y[, 2L] * 0.8, col = col[2L], length =
## arrow.len): zero-length arrow is of indeterminate angle and so skipped
## Warning in arrows(0, 0, y[, 1L] * 0.8, y[, 2L] * 0.8, col = col[2L], length =
## arrow.len): zero-length arrow is of indeterminate angle and so skipped
Repeated with standardized data and variance percentages to the labels.
human_std <- scale(human)
pca_human_std <- prcomp(human_std)
s <- summary(pca_human_std)
pca_pr <- round(100*s$importance[2, ], digits = 1)
pc_lab <- paste0(names(pca_pr), " (", pca_pr, "%)")
biplot(pca_human_std, cex = c(0.8, 1), col = c("grey40", "deeppink2"), xlab = pc_lab[1], ylab = pc_lab[2])
Plots side-by-side
par(mfrow = c(1,2))
biplot(pca_human, cex = c(0.8, 1), col = c("grey40", "deeppink2"), xlab = pc_lab[1], ylab = pc_lab[2])
## Warning in arrows(0, 0, y[, 1L] * 0.8, y[, 2L] * 0.8, col = col[2L], length =
## arrow.len): zero-length arrow is of indeterminate angle and so skipped
## Warning in arrows(0, 0, y[, 1L] * 0.8, y[, 2L] * 0.8, col = col[2L], length =
## arrow.len): zero-length arrow is of indeterminate angle and so skipped
## Warning in arrows(0, 0, y[, 1L] * 0.8, y[, 2L] * 0.8, col = col[2L], length =
## arrow.len): zero-length arrow is of indeterminate angle and so skipped
## Warning in arrows(0, 0, y[, 1L] * 0.8, y[, 2L] * 0.8, col = col[2L], length =
## arrow.len): zero-length arrow is of indeterminate angle and so skipped
## Warning in arrows(0, 0, y[, 1L] * 0.8, y[, 2L] * 0.8, col = col[2L], length =
## arrow.len): zero-length arrow is of indeterminate angle and so skipped
biplot(pca_human_std, cex = c(0.8, 1), col = c("grey40", "deeppink2"), xlab = pc_lab[1], ylab = pc_lab[2])
MCA is kinda like PCA, but for categorical data.
Access the FactoMineR library and the tea dataset.
library(FactoMineR)
data(tea)
Select only the instructed columns and check summary and structure.
keep_columns <- c("Tea", "How", "how", "sugar", "where", "lunch")
tea_time <- select(tea, one_of(keep_columns))
summary(tea_time)
## Tea How how sugar
## black : 74 alone:195 tea bag :170 No.sugar:155
## Earl Grey:193 lemon: 33 tea bag+unpackaged: 94 sugar :145
## green : 33 milk : 63 unpackaged : 36
## other: 9
## where lunch
## chain store :192 lunch : 44
## chain store+tea shop: 78 Not.lunch:256
## tea shop : 30
##
str(tea_time)
## 'data.frame': 300 obs. of 6 variables:
## $ Tea : Factor w/ 3 levels "black","Earl Grey",..: 1 1 2 2 2 2 2 1 2 1 ...
## $ How : Factor w/ 4 levels "alone","lemon",..: 1 3 1 1 1 1 1 3 3 1 ...
## $ how : Factor w/ 3 levels "tea bag","tea bag+unpackaged",..: 1 1 1 1 1 1 1 1 2 2 ...
## $ sugar: Factor w/ 2 levels "No.sugar","sugar": 2 1 1 2 1 1 1 1 1 1 ...
## $ where: Factor w/ 3 levels "chain store",..: 1 1 1 1 1 1 1 1 2 2 ...
## $ lunch: Factor w/ 2 levels "lunch","Not.lunch": 2 2 2 2 2 2 2 2 2 2 ...
Bar plots
library(ggplot2)
p <- gather(tea_time) %>% ggplot(aes(value)) + facet_wrap("key", scales = "free")
p <- p + geom_bar()
p + theme(axis.text.x = element_text(angle = 45, hjust = 1, size = 8))
Produce the MCA model and print the summary
mca <- MCA(tea_time, graph = FALSE)
summary(mca)
##
## Call:
## MCA(X = tea_time, graph = FALSE)
##
##
## Eigenvalues
## Dim.1 Dim.2 Dim.3 Dim.4 Dim.5 Dim.6 Dim.7
## Variance 0.279 0.261 0.219 0.189 0.177 0.156 0.144
## % of var. 15.238 14.232 11.964 10.333 9.667 8.519 7.841
## Cumulative % of var. 15.238 29.471 41.435 51.768 61.434 69.953 77.794
## Dim.8 Dim.9 Dim.10 Dim.11
## Variance 0.141 0.117 0.087 0.062
## % of var. 7.705 6.392 4.724 3.385
## Cumulative % of var. 85.500 91.891 96.615 100.000
##
## Individuals (the 10 first)
## Dim.1 ctr cos2 Dim.2 ctr cos2 Dim.3
## 1 | -0.298 0.106 0.086 | -0.328 0.137 0.105 | -0.327
## 2 | -0.237 0.067 0.036 | -0.136 0.024 0.012 | -0.695
## 3 | -0.369 0.162 0.231 | -0.300 0.115 0.153 | -0.202
## 4 | -0.530 0.335 0.460 | -0.318 0.129 0.166 | 0.211
## 5 | -0.369 0.162 0.231 | -0.300 0.115 0.153 | -0.202
## 6 | -0.369 0.162 0.231 | -0.300 0.115 0.153 | -0.202
## 7 | -0.369 0.162 0.231 | -0.300 0.115 0.153 | -0.202
## 8 | -0.237 0.067 0.036 | -0.136 0.024 0.012 | -0.695
## 9 | 0.143 0.024 0.012 | 0.871 0.969 0.435 | -0.067
## 10 | 0.476 0.271 0.140 | 0.687 0.604 0.291 | -0.650
## ctr cos2
## 1 0.163 0.104 |
## 2 0.735 0.314 |
## 3 0.062 0.069 |
## 4 0.068 0.073 |
## 5 0.062 0.069 |
## 6 0.062 0.069 |
## 7 0.062 0.069 |
## 8 0.735 0.314 |
## 9 0.007 0.003 |
## 10 0.643 0.261 |
##
## Categories (the 10 first)
## Dim.1 ctr cos2 v.test Dim.2 ctr cos2
## black | 0.473 3.288 0.073 4.677 | 0.094 0.139 0.003
## Earl Grey | -0.264 2.680 0.126 -6.137 | 0.123 0.626 0.027
## green | 0.486 1.547 0.029 2.952 | -0.933 6.111 0.107
## alone | -0.018 0.012 0.001 -0.418 | -0.262 2.841 0.127
## lemon | 0.669 2.938 0.055 4.068 | 0.531 1.979 0.035
## milk | -0.337 1.420 0.030 -3.002 | 0.272 0.990 0.020
## other | 0.288 0.148 0.003 0.876 | 1.820 6.347 0.102
## tea bag | -0.608 12.499 0.483 -12.023 | -0.351 4.459 0.161
## tea bag+unpackaged | 0.350 2.289 0.056 4.088 | 1.024 20.968 0.478
## unpackaged | 1.958 27.432 0.523 12.499 | -1.015 7.898 0.141
## v.test Dim.3 ctr cos2 v.test
## black 0.929 | -1.081 21.888 0.382 -10.692 |
## Earl Grey 2.867 | 0.433 9.160 0.338 10.053 |
## green -5.669 | -0.108 0.098 0.001 -0.659 |
## alone -6.164 | -0.113 0.627 0.024 -2.655 |
## lemon 3.226 | 1.329 14.771 0.218 8.081 |
## milk 2.422 | 0.013 0.003 0.000 0.116 |
## other 5.534 | -2.524 14.526 0.197 -7.676 |
## tea bag -6.941 | -0.065 0.183 0.006 -1.287 |
## tea bag+unpackaged 11.956 | 0.019 0.009 0.000 0.226 |
## unpackaged -6.482 | 0.257 0.602 0.009 1.640 |
##
## Categorical variables (eta2)
## Dim.1 Dim.2 Dim.3
## Tea | 0.126 0.108 0.410 |
## How | 0.076 0.190 0.394 |
## how | 0.708 0.522 0.010 |
## sugar | 0.065 0.001 0.336 |
## where | 0.702 0.681 0.055 |
## lunch | 0.000 0.064 0.111 |
visualize MCA
plot(mca, invisible=c("ind"), habillage = "quali")